Question: Evaluate $\int\sin^2x\cos^3x\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\sin^3x}3-\dfrac{\sin^5x}5+C$ (Choice B) B $\dfrac{\sin^3x}3-\dfrac{\sin^4x}4+C$ (Choice C) C $\dfrac{\cos^3x}3-\dfrac{\cos^5x}5+C$ (Choice D) D $\dfrac{\cos^3x}3-\dfrac{\cos^4x}4+C$
Solution: This has an odd power of $~\cos x\,$. We factor out a single factor of $~\cos x~$ and use the identity $~\cos^2x=1-\sin^2x~$ to rewrite everything else in terms of $~\sin x\,$. $\begin{aligned} \int\sin^2x\cos^3x\,dx&=\int \sin^2x\cos^2x\cos x\,dx\\ \\&=\int\sin^2x(1-\sin^2x)\cos x\,dx\\ \\&=\int(\sin^2x-\sin^4)\cos x\,dx\end{aligned}$ We can now use a $~u$ -substitution where $~u=\sin x~$ and $~du=\cos x\,dx\,$. $ \int(\sin^2x-\sin^4)\cos x\,dx~~~~~\Rightarrow~~~~~\int (u^2-u^4)\,du$ Antidifferentiate. $ \int (u^2-u^4)\,du=\dfrac{u^3}3-\dfrac{u^5}5+C$ Now substitute back to obtain the result in terms of $~x\,$. $ \dfrac{u^3}3-\dfrac{u^5}5+C~~~~~\Rightarrow~~~~~\int\sin^2x\cos^3x\,dx=\dfrac{\sin^3x}3-\dfrac{\sin^5x}5+C$